A smooth bead or ring with non-zero mass is threaded onto a light inextensible string that is tied at each end to hooks A and B on the ceiling of a room.

A force is applied to the ring such that the entire system is held in equilibrium, with each segment of string taut.

In such a scenario, is it correct to assume that the tension in both segments of string is the same due to the fact that the ring or bead is modeled as being smooth?

Thank you in advance for clarification.

If the system is in free equilibrium, I.e the ring is being held by the string alone and isn't fixed into a certain position then the tension will be equal. If the ring is being held in position away from its natural position then the tensions wouldn't be equal...
Hope this helps

Does anyone disagree with ronrabbit's post?

By virtue of their solutions / mark schemes, the exam questions I'm looking at all assume the tension is constant in each string segment (even when a force is applied to the ring to hold the system in equilibrium).

Thanks again.

Smoothness only means that you don't have to worry about friction of the hooks in your calculation.

It is true that each segment of the string will have the same tension (consider the force diagram at some point in the middle of the string -- you can't have a difference otherwise there's a net force). But it is not true that the entire string must have the same tension throughout.

Where you intuition might be leading you wrong is that you're mistaking the lack of friction between the string and the hook as meaning that the tension "passes through" the hook in some way (that the tension "curves around" the hook as if the hook isn't there). But that's not what's happening.

You want to think of the string and hook touching at a single point. If you cut that string at that exact point and then reconnect the strings to the hook there, you should see the exact same force diagram, but now you have two separate strings. And so all the calculations are the same for the connected string case as the two-string case.

Drawing or it didnt happen lol!

I've gone over this in my head again, and it doesn't seem as right as it did when I first typed it up. You end up with an underconstrained system if you do this. I just read ronrabbit's response, and he's got the constraint that I was missing.

Depends on the direction of the force.... if it is exactly in the middle I would say yes...offset I would say one side would be more loaded than the other......also a 10th grade dropout so don't bet the farm

Any cylindrically symmetric configuration (not accelerating or rotating) along the gravitational axis will have equal forces. But a drawing with details, because nothing is perfectly symmetric or uniform in density, helps. Eg it can look symmetric and sill have center of mass outside the obvious geometric assumed one.

For solid bodies that are not point particles you need to do both a forces and torques equilibrium analysis.

Drawing or it didnt happen is a joke but one that is to be taken seriously. A good drawing or picture always inspires good ideas and promotes accuracy.